Mark Kozek. December 7, 2010

Transcription

1 : in : Whittier College December 7, 2010

2 About. : in Hungarian mathematician, Interested in combinatorics, graph theory, number theory, classical analysis, approximation theory, set theory, probability theory...

3 More on. : in Wrote/co-wrote over 1400 papers. Erdős number. s: issued bounties for problems that he thought were interesting or for which he wanted to know the solution. Erdős resources: 1 The man who loved only numbers, book by Paul Hoffman, 1998, (based on his Atlantic Monthly article from 1989). 2 N is a Number: A portrait of, film by George Csicsery, 1993 (also based on the Atlantic Monthly article). 3 And what is your Erdős number?", Caspar Goffman, the American Mathematical Monthly, 1969.

4 Erdős in : in Question: (de Polignac, 1849) Is it the case that every sufficiently large odd integer > 1 can be written as the sum of a prime number and a power of 2? Some small counter examples include: 127, 905. (Romanoff) A positive proportion of the integers may be expressed this way. (van der Corput) The exceptions form a set of positive density. (Erdős) Constructed an arithmetic progression of odd integers not representable in this way.

5 Background: (from Number Theory 101) : Def.: Congruent The integers a and b are congruent modulo the natural number n > 1 if there exists an integer, z such that a b = zn. in If so, we write For example, a b (mod n) (mod 10) (mod 1200) 3 58 (mod 5) (mod 11).

6 Background: (from Number Theory 101) : in Def.: system A covering system or covering, for short, is a finite system of congruences n a i (mod m i ), 1 < i t, such that every integer satisfies at least one of the congruences.

7 Example of a covering. : in For example, the congruences n 0 (mod 2) n 1 (mod 3) n 3 (mod 4) n 5 (mod 6) n 9 (mod 12) form a covering of the integers.

8 More background. : in Def.: Chinese Remainder Theorem A system of congruences has a unique solution if the moduli are pairwise relatively prime. For example: Can we solve for x in the following system? x 3 mod 9 x 5 mod 10 x 2 mod 11 Yes, because {9,10,11} are pair-wise, relatively prime. We get: (Note: 990 = ) x 255 mod 990

9 Back to de Polignac s question. : in Question: (de Polignac, 1849) Is it the case that every sufficiently large odd integer > 1 can be written as the sum of a prime number and a power of 2? de Polignac asked about odd numbers of the form x = p + 2 n. Erdős instead thought about writing primes in the form p = x 2 n, and changed the question to: Question : (Erdős) Is it possible to find an integer x such that x 2 n is not prime for all (non-negative) integers n?

10 Erdős approach to de Polignac s conjecture : in This led to the sub-question : For which n and for which x would x 2 n be divisible by 3? x 2 n = 3z x 2 n 0 (mod 3) x 2 n (mod 3) Let us take powers of 2 (mod 3) We see that: 2 n 0 (mod 3) (mod 3) (mod 3) (mod 3) (mod 3) For even powers of 2, 2 n 1 (mod 3). For odd powers of 2, 2 n 2 (mod 3).

11 Erdős approach to de Polignac s conjecture : in This led to the sub-question : For which n and for which x would x 2 n be divisible by 3? We have that if, n 0 (mod 2) (even) and x 1 (mod 3) then x 2 n is divisible by 3. OR n 1 (mod 2) (odd) and x 2 (mod 3) then x 2 n is divisible by 3.

12 Erdős approach to de Polignac s conjecture : in Erdős strategy was to continue along these lines and try to find conditions on n and on x that would ensure that x 2 n would be divisible by primes from a given set (that would include 3). He found the following relations: n 0 (mod 2) and x 1 (mod 3) 3 x 2 n n 0 (mod 3) and x 1 (mod 7) 7 x 2 n n 1 (mod 4) and x 2 (mod 5) 5 x 2 n n 3 (mod 8) and x 8 (mod 17) 17 x 2 n n 7 (mod 12) and x 11 (mod 13) 13 x 2 n n 23 (mod 4) and x 121 (mod 241) 241 x 2 n

13 Erdős approach to de Polignac s conjecture : in n 0 (mod 2) and x 1 (mod 3) 3 x 2 n n 0 (mod 3) and x 1 (mod 7) 7 x 2 n n 1 (mod 4) and x 2 (mod 5) 5 x 2 n n 3 (mod 8) and x 8 (mod 17) 17 x 2 n n 7 (mod 12) and x 11 (mod 13) 13 x 2 n n 23 (mod 24) and x 121 (mod 241) 241 x 2 n We observe that the set of congruences describing n form a covering system. We observe that the set of congruences describing x can be combined solved using Chinese Remainder Theorem.

14 Erdős arithmetic progression of counterexamples : in By the Chinese Remainder Theorem, we get where x (mod ), = This gives us the Erdős arithmetic progression of counterexamples to de Polignac s conjecture. k = ± z, for z Z.

15 Erdős arithmetic progression of counterexamples : in There exists an arithmetic progression of odd integers x, that simultaneously satisfy the system of congruences x 1 (mod 3) x 1 (mod 7) x 2 (mod 5) x 8 (mod 17) x 11 (mod 13) x 121 (mod 241) x 1 (mod 2), such that x 2 n is composite (not prime) for all non-negative integers n because x 2 n will be divisible by at least one of the primes from the set {3, 5, 7, 13, 17, 241}.

16 Question: What next? : in Answer: We try to generalize. Recall, Erdős constructed an arithmetic progression of odd integers x such that x 2 n was composite for all non-negative integers n. Def.: Sierpiński number A Sierpiński number is a positive odd integer k with the property that k 2 n + 1 is composite for all positive integers n.

17 Sierpiński numbers : in Def.: Sierpiński number A Sierpiński number is a positive odd integer k with the property that k 2 n + 1 is composite for all positive integers n. Sierpiński (1960) observed the following implications: n 1 (mod 2), k 1 (mod 3) = k 2 n (mod 3) n 2 (mod 4), k 1 (mod 5) = k 2 n (mod 5) n 4 (mod 8), k 1 (mod 17) = k 2 n (mod 17) n 8 (mod 16), k 1 (mod 257) = k 2 n (mod 257) n 16 (mod 32), k 1 (mod 65537) = k 2 n (mod 65537) n 32 (mod 64), k 1 (mod 641) = k 2 n (mod 641) n 0 (mod 64), k 1 (mod ) = k 2 n (mod ).

18 Sierpiński numbers : in The moduli appearing in the congruences involving k are 7 primes, the first (perhaps only) five Fermat primes F n = 2 2n + 1 for 0 n 4 and the two prime divisors of F 5. We add the condition k 1 (mod 2) to ensure that k is odd. Then the Chinese Remainder Theorem implies that there are infinitely many Sierpiński numbers given by k (mod ).

19 Smallest Sierpiński number : in In 1962, Selfridge (unpublished) found what is believed to be the smallest Sierpiński number, namely k = His argument is based on the following implications: n 0 (mod 2), k 2 (mod 3) = k 2 n (mod 3) n 1 (mod 4), k 2 (mod 5) = k 2 n (mod 5) n 3 (mod 9), k 9 (mod 73) = k 2 n (mod 73) n 15 (mod 18), k 11 (mod 19) = k 2 n (mod 19) n 27 (mod 36), k 6 (mod 37) = k 2 n (mod 37) n 1 (mod 3), k 3 (mod 7) = k 2 n (mod 7) n 11 (mod 12), k 11 (mod 13) = k 2 n (mod 13). There have been attempts to prove that is the smallest Sierpiński number. In this regard, the web page contains the current up-to-date information.

20 Smallest Sierpiński number : in As of this writing, there remain 6 values of k < which are unresolved by the Seventeen or Bust project, namely 10223, 21181, 22699, 24737, 55459, The most recent value of k < to have been eliminated was 33661, by Sturle Sunde s computer, on October 30, 2007.

21 Other generalizations. : in Def.: number A number is a positive odd integer k with the property that k 2 n 1 is composite for all positive integers n. The smallest known number is , due to (1956). There have been attempts to prove that is the smallest number. As of this writing there remain 64 unresolved candidates, of these 2293 is the smallest.

22 More generalizations : in Conjecture (Chen) For every positive integer r, there exist infinitely many positive odd integers k such that the number k r 2 n + 1 has at least two distinct prime factors for each positive integer n. Conjecture (Chen) For every positive integer r, there exist infinitely many positive odd integers k such that the number k r 2 n has at least two distinct prime factors for each positive integer n. (Equivalent to k r 2 n 1.)

23 Chen s conjectures. : in Chen (2002) resolves each conjecture in the case that r is odd and in the case that r is twice an odd number and 3 r. As he notes, the least r for which his arguments do not apply are r = 4 and r = 6. Conjecture 1 is true in general and that Conjecture 2 holds in the special cases r = 4 and r = 6. Theorem (Filaseta, Finch, K., 2008) For every positive integer R, there exist infinitely many positive odd numbers k such that each of the numbers k2 n + 1, k 2 2 n + 1, k 3 2 n + 1,..., k R 2 n + 1 has at least two distinct prime factors for each positive integer n.

24 The minimum modulus problem : in Open problem: (Erdős, $1000) For every natural number N > 1 does there exist a covering system with distinct moduli all N? Personal attempts: 1 Summer 2008, with Kelly Bickell, Michael Firrisa, Juan Pablo Ortiz, and Kristen Pueschel. We made it to N = Summer 2009, with Tobit Raff. We made it to N = 11.

25 The minimum modulus problem: results : in Open problem: (Erdős, $1000) For every natural number N > 1 does there exist a covering system with distinct moduli all N? Early results: N = 3: Erdős. N = 9: Churchhouse, N = 14: Selfridge. N = 18: Krukenburg (Ph.D. Thesis), N = 20: Choi, N = 24: Morikawa, N = 25: Gibson (Ph.D. Thesis) N = 36, 40: Nielsen, 2009.

26 The minimum modulus problem: techniques : in For small N, examples can be worked out by hand, but quickly computers come into play. Churchouse s result (N=9) came from using computers and a greedy algorithm. The LCM of the moduli was 604, 800 = Krukenburg and Choi s results did not use computers. The LCM of Krukenburg s moduli was 475, 371, 719, 222, 400 =

27 Gibson s techniques : in Gibson uses: a greedy algorithm (like Churchhouse) the notion of an almost covering (like Morikawa who in turned used ideas of Krukenburg) random covering (like Erdős) extensive computing. The LCM of the moduli used primes up to 2017.

28 Nielsen s techniques : in Nielsen uses a graph theoretic approach, representing covering as trees, and introducing new primes, as necessary to plug holes. The LCM of the moduli uses primes up to 103. Initially, he used the primes in order. However, the referee noted that sometimes it was more efficient to use certain primes out of order. This allowed for the improvement from N = 36 to N = 40. There was very little wiggle room, and thus, fears a negative solution for the minimum modulus problem.

29 Sierpiński +? : in Question: Do here exist numbers that are simultaneously numbers? Answer: Yes. (Cohen and Selfridge, 1975). k =

30 Sierpiński +? : in However, folks (computer scientists) didn t read their paper. So they offered their results. Brier (1998) k = Gallot (2000) k = Gallot (2000) k = Gallot (2000) k = E. Vantieghem (2010) k = Filaseta, Finch and K. (2008) k = For more information on this problem, visit:

31 Thank you : in Any Questions?

32 In Memoriam : in John L. Selfridge